Question

If $3x-4y+k=0$ is a tangent to the hyperbola $x^2-4y^2=5$ find the value of k.

Answer 1

The equation of the hyperbola is $x^2-4y^2=5$

$\frac{x^2}{5}-\frac{y^2}{\left(\frac{5}{4}\right)}=1$

This is of the form, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$⟹a^2=5$ and $b^2=\frac{5}{4}$

Equation of the line is $3x-4y+k=0$

$⟹y=\frac{3}{4}x+\frac{k}{4}$ ------ (i)

This is of the form, $y=mx+c$, the equation of a straight line, with $m=\frac{3}{4}$ and $c=\frac{k}{4}$

If (i) is a tangent to the hyperbola, then,

$c^2=a^2{m}^2-b^2$

$\frac{k^2}{16}=5.\frac{9}{16}-\frac{5}{4}$

$k^2=45-20$

$k^2=25$

$k=\pm 5$